$6!\cdot 7!=10!$. Is there a natural bijection between $S_6\times S_7$ and $S_{10}$?

Question

Aside from $1!\cdot n!=n!$ and $(n!-1)!\cdot n! = (n!)!$, the only product of factorials known is $6!\cdot 7!=10!$.

One might naturally associate these numbers with the permutations on $6, 7,$ and $10$ objects, respectively, and hope that this result has some kind of connection to a sporadic relation between such permutations - numerical "coincidences" often have deep math behind them, like how $1^2+2^2+\ldots+24^2=70^2$ can be viewed as an ingredient that makes the Leech lattice work.

The most natural thing to hope for would be a product structure on the groups $S_6$ and $S_7$ mapping to $S_{10}$, but as this MathOverflow thread shows, one cannot find disjoint copies of $S_6$ and $S_7$ living in $S_{10}$, so a product structure seems unlikely.

However, I'm holding out hope that some weaker kind of bijection can be found in a "natural" way. Obviously one can exhibit a bijection. For instance, identify the relative ordering of $1,2,\ldots 7$ in a permutation of size $10$, and then biject $_{10}P_{3}=720$ with $S_6$ in some way. But I'd like to know if there is a way to define such a bijection which arises naturally from the permutation structures on these sets, and makes it clear why the construction does not extend to other orders.

I tried doing something with orderings on polar axes of the dodecahedron ($10!$) and orderings on polar axes of the icosahedron ($6!$), in the hopes that the sporadic structure and symmetry of these Platonic solids would allow for interesting constructions that don't generalize, but ran into issues with the dodecahedron (sequences of dodecahedral axes aren't particularly nice objects) and the question of how to extract a permutation of length $7$.

I'm curious if someone can either devise a natural bijection between these sets or link to previous work on this question.

Answer 1

This family of bijections (of sets) $S_6\times S_7 \to S_{10}$ has already been suggested in comments and linked threads, but it is so pretty I wanted to spell it out:

There are $10$ ways of partitioning the numbers $1,2,3,4,5,6$ into two (unordered) pieces of equal size: $P_1,P_2,\cdots,P_{10}$. Thus we have a canonical embedding $S_6\hookrightarrow S_{10}$, coming from the induced action on the $P_i$.

Any distinct pair $P_i,P_j$ will be related by a unique transposition. For example $\{\{1,2,3\},\{4,5,6\}\}$ (denoted hereafter $\left(\frac{123}{456}\right)$) is related to $\left(\frac{126}{453}\right)$ via the transposition $(36)$.

There are two types of ordered (distinct) triples $P_i, P_j,P_k$:

1. They may be related pairwise via transpositions $(ab),(cd),(ef)$ with $a,b,c,d,e,f$ distinct and each of $\{a,b\}, \{c,d\},\{e,f\}$ not on the same side of any of $P_i, P_j,P_k$:$$ \left(\frac{ace}{bdf}\right), \left(\frac{bce}{adf}\right), \left(\frac{ade}{bcf}\right).$$
   Here, there are $10$ choices for $P_i$, $9$ choices for $P_j$ and $4$ choices for $P_k$, giving $360$ triples in total.

2. They may be related pairwise via transpositions $(ab),(ca),(bc)$ with $a,b,c$ distinct: $$ \left(\frac{ace}{bdf}\right), \left(\frac{bce}{adf}\right), \left(\frac{abe}{cdf}\right).$$
   Again, there are $10$ choices for $P_i$, $9$ choices for $P_j$ and $4$ choices for $P_k$, giving $360$ triples in total.

An element of the stabiliser (in $S_6$) of a type 1 ordered triple (written as above) must preserve the pairs $\{a,b\}, \{c,d\},\{e,f\}$. Further if it swaps any of these pairs it must swap all of them, so the only non-trivial element of the stabiliser is an odd permutation: $(ab)(cd)(ef)$.

An element of the stabiliser (in $S_6$) of a type 2 ordered triple (written as above) must preserve the sets $\{d,f\}, \{e\},\{a,c,b\}$. Further it must fix each of $a,b,c$. Thus the only non-trivial element of the stabiliser is an odd permutation: $(df)$.

As $|A_6|=360$, in particular this means there is a unique element of $A_6$ taking the ordered triple $P_1,P_2,P_3$ to a specified ordered triple $P_i,P_j,P_k$ of the same type as $P_1,P_2,P_3$.

Fix $t\in S_{10}$ a permutation taking $P_1,P_2,P_3$ to an ordered triple of the other type. Then there is a unique element in $A_6$ which composed with $t$ takes the ordered triple $P_1,P_2,P_3$ to a specified ordered triple $P_i,P_j,P_k$ of the other type to $P_1,P_2,P_3$.

Let $S_7$ denote the group of permutations of $P_4,P_5,\cdots,P_{10}$. Then any permutation in $S_{10}$ may be written uniquely as an element of $S_7$ followed by an element of $(A_6\sqcup tA_6)$, where the latter is determined by where $P_1,P_2,P_3$ are mapped to.

Thus we have established a bijection of sets $$S_{10}\to (A_6\sqcup tA_6)\times S_7.$$ Once we fix an odd permutation $t'\in S_6$, we may identify the sets $$(A_6\sqcup t'A_6)\to S_6.$$ Composing we get: $$S_{10}\to (A_6\sqcup tA_6)\times S_7\to (A_6\sqcup t'A_6)\times S_7\to S_6\times S_7.$$

That is for any choice of the permutations $t,t'$ we have the required bijection of sets.

Answer 2

It may be connected with, of all things, the $3-4-5$ right triangle! This triangle and its multiples stand out as having the sides in arithmetic progression. Such an arithmetic progression leads to factorial expressions when the sides are multiplied together.

As a preliminary step, consider a relatively unheralded property of right triangles: the diameter of the incircle plus the hypotenuse equals the sum of the other two sides. Suppose that the legs are $a$ and $b$, and the hypoteneuse is $c$ where $c^2=a^2+b^2$. The diameter of the incircle is then $2ab/(a+b+c)$ while the Pythagorean relation implies $$(a+b+c)(a+b-c)=(a^2+2ab+b^2)-(a^2+b^2)=2ab$$ Thereby the diameter of the incircle reduces to $a+b-c$. Should there be a right triangle whose sides are in arithmetic progression, then, the diameter of the incircle will join this progression, making it longer and thus perhaps generating a bigger factorial upon multiplication.

In this question it is shown that the product of the sides of any triangle is half the product of the diameter of the circumcircle (circumdiameter), the diameter of the incircle (indiameter), and the perimeter. Let us see where that leads if we apply it to a right triangle having sides $3,4,5$. Multiplying the sides together then gives

$3×4×5=\text{circumdiameter}×\text{indiameter}×\text{perimeter}/2$

We double the sides of the triangle to clear the fraction on the right side:

$6×8×10=\text{circumdiameter}×\text{indiameter}×\text{perimeter}×4$

The circumdiameter is the hypoteneuse of the $3-4-5$ triangle, thus $5$ -- which is in the aforementioned arithmetic progression. The indiameter is $2$ from the above lemma, which precedes $3,4,5$ in the arithmetic progression. And the perimeter of the triangle is three times the longer leg, again due to the arithmetic progression, thus $4×3$. Substituting these results into the above product equality then gives

$6×8×10=5×2×(3×4)×4=5!×4$

And there is our factorial. To make it cleaner we should multiply by $3/2$, absorbing the dangling factor $4$ into the factorial. We then get three different three-term products on the left side, depending on which of the factors $6,8,10$ we increment:

$\color{blue}{8×9×10}=6×10×12=6×8×15=5×2×(3×4)×6=6!$

And from the three-term product shown in blue, we have

$6!=10!/7!$

Why is this uniquely chosen? We see that the sides of a right triangle being in arithmetic progression lead to the factorial on the right in two ways, by making the perimeter a simple multiple of one leg and by incorporating the circum diameter into the arithmetic progression. Only the $3-4-5$ right triangle has these properties, and it leads specifically to $6!$ also being a factorial ratio.